Question

Chapter 5, Reserve Problem 078 A steel [E = 29,100 ksi; a =6.4 x 10-6/0F] pipe column (1) with a cross-sectional area of A4 = 5.20 in.2 is connected at flange B to an aluminum alloy [E = 10,200 ksi; a =12.8 x 10-6/0F] pipe (2) with a cross-sectional area of Az = 4.10 in.2. The assembly (shown in the figure) is connected to rigid supports at A and C. It is initially unstressed at a temperature of 93°F. Assume L1 = 143 in., L2 = 172 in., P = 34 kips. (a) At what temperature will the normal stress in steel pipe (1) be reduced to zero? (b) Determine the normal stresses (positive if tensile, negative if compressive) in steel pipe (1) and aluminum pipe (2) when the temperature reaches -12°F. (1) (2) с A B Р L L2 Answers: (a) T = OF (b) 01 = ksi 02 = ksi

Answer 1

F F2 B FBD a) Equilibrium: consider a FBD of flange B. sum form forces in the horizontal direction to obtain equation (1). F =-F, + F, - 68 kips =0 Geometry of deformations to obtain equation (2) & + o2 = 0 Force- temperature-deformation relationship give equation (3) & = FL +ą,474 AE FL2 +a,AT,L AEZ 82 = Compatibility equation: Substitute eq (3) into eq. (2) to derive the compatibility equation to get equation (4).

FL FL2 +QATL + +AT,L2 = 0 AE A,E2 Solve the equations: Set F1 = 0 and solve eq. (1) to find F, = 68kips. substitute these values for F, and F, into eq. (4) along with the observation that the temperature change for both axial members is the same (i.e.,AT, = AT, = AT) and solve for AT : FL, F,L2 4E, AE, AT = aL, +aL (68 kips) (172 in.) 0- (4.10 in?) (10,200ksi) AT 6.4x10-6 12.8x10-6 (143 in.) + (172in.) F F 0.279675 3.1168 x 10 AT = -89.73°F = Since the pipes are initially at a temperature 93°F the temperature at which the normal stress in steel pipe (1) is reduced to zero is T = 93°F -89.73F T = 3.2685°F

b) Solve eq.(1) for F, to obtain equation (5) F2 = F+68 kips When the temperature reaches –12°F the total change in temperature is AT =-(93+12)°F =-105°F substitute this value along with eq. (5) into the compatibility equation (eq. (4)] and derive an expression for F: + + + FL (F. +68kips) :-QATL - ATL AE A₂E₂ L L (68 kips) L2 F --AT[a,L, +a,L] 4E, A2E2 AE₂ L L (68 kips), F =-AT [Q, L, +a,L]- 4 E, A,E2 (68 kips) L2 -AT[04 +aL2]- AE F = L. L AE, A,E + A E2 + And compute F, :

-(-105°F)[(6.4*10^)(143in.)+(12.8–10“)(172 in )]- (4.102 )(10,200ksi) (68kips) (172 in) F= II 143 in 172 in (5.20in.? )(29,000 ksi) (4.10in.? )(10,200 ksi) + 0.327264 in -0.27967 in. F = in in. 9.482758 x 10-4 +41.1286x10-4 kip kip 0.047594 in. F in 5.061x10- kip F = 9.404 kips = From eq. (a), F, has a value of F = F, +68kips = 9.404 kips +68kips = 77.4kips Normal stresses:

The normal stresses in each axial member can be calculated by following: F A 9.404 kips 0 5.20 in. 0=1.808ksi 0 =1.808 ksi(T F, 02 = 77.4 kips 02 4.10 in. 02=18.88ksi 02=18.88ksi(T)

The answers are boxed.